Having examined the various aspects that can influence the design of steel beams, the general procedure when using grade 43 rolled sections may be summarized as follows.
Bending
(a) Decide if the beam will be laterally restrained or laterally unrestrained.
(b) If the beam is laterally restrained, ensure that the moment capacity Mcx of the section is greater than the applied ultimate moment Mu:
Mcx = pySx≥ Mu
(c) If the beam is laterally unrestrained, the lateral torsional buckling resistance of the section will have to be checked. This may be done using a rigorous approach or a conservative approach. In both methods, account should be taken of any loading between restraints.
(i) Using the rigorous approach, ensure that the applied equivalent uniform moment M is less than the buckling resistance moment Mb of the section:
M = mMA ≤ Mb = pbSx
(ii) Using the conservative approach, ensure that the maximum mo-ment Mx occurring between lateral restraints does not exceed the buckling resistance moment Mb of the section:
Mx ≤ Mb = pbSx
Shear
Both the combination of maximum moment and coexistent shear, and the combination of maximum shear and coexistent moment, should be checked.
The shear resistance of a beam is checked by ensuring that the ultimate shear force Fv does not exceed the shear capacity Pv of the section at the point under consideration:
Fv≤ Pv
It should be noted that the moment capacity of plastic and compact beam sections must be reduced when high shear loads occur. However, this is not usually a problem except for heavily loaded short span beams.
A high shear load condition exists when Fv > 0.6 Pv
Deflection
The deflection requirement of a beam is checked by comparing the actual deflection produced by the unfactored imposed loads with the recom-mended limits given in BS 5950 Table 5:
Actual deflection < recommended deflection limit
Web buckling
The web buckling resistance of an unstiffened web must be greater than any concentrated load that may be applied.
Web bearing
The web bearing resistance of an unstiffened web must be greater than any concentrated load that may be applied.
It should be appreciated that the requirements for web buckling and bearing are not usually critical under normal loading conditions. Further-more, they can if necessary be catered for by the inclusion of suitably designed web stiffeners.
Before leaving the topic of beams, let us look at a further example illustrat-ing the complete design of a laterally unrestrained beam usillustrat-ing the rigorous approach.
W1 W 2
A B C
3 m 3 m 3 m D
A
Ra = 112.97 kN
Example 5.10
The simply supported beam shown in Figure 5.24 is laterally restrained at the ends and at the points of load application. For the loads given below, determine the size of grade 43 section required.
9 m
Figure 5.24 Simply supported beam Specified dead loads:
Point load W1d = 30 kN; point load W2d = 20 kN Self-weight = 1 kN/m; SW UDL = 1 × 9 = 9 kN Specified imposed loads:
Point load W1i = 50 kN; point load W2i = 30 kN Ultimate design loads:
W1 = f W1d + f W1i = 1.4 × 30 + 1.6 × 50 = 42 + 80 = 122 kN W2 = f W2d + f W2i = 1.4 × 20 + 1.6 × 30 = 28 + 48 = 76 kN SW UDL = 1.4 × 9 = 12.6 kN
The ultimate design load diagram and the corresponding shear force and bending moment diagrams are shown in Figure 5.25. Since the loading is not symmetrical, the reactions and moments are calculated from first principles.
122 kN 76 kN
Self weight UDL = 12.6 kN
B C D
3 m 3 m 3 m
9 m Rd = 97.63 kN
112.97 kN
108.77 kN +
13.23 kN 17.43 kN –
93.43 kN 97.63 kN (a) Ultimate design load diagram (b) Shear force diagram
332.61 kNm 286.59 kNm
(c) Bending moment diagram
Figure 5.25 Beam diagrams for ultimate loads
+
γ γ
γ γ
For the reactions, take moments about D:
9 Ra = (122 × 6) + (76 × 3) + (12.6 × 4.5) = 732 + 228 + 56.7 = 1016.7 Ra = 1016.7/9 = 112.97 kN
Rd = (122 + 76 + 12.6) – 112.97 = 97.63 kN 12.6 32
Ultimate moment at B = 112.97 × 3 – × = 332.61 kN m 9 2
12.6 32
Ultimate moment at C = 97.63 × 3 – × = 286.59 kN m 9 2
Since the beam is not fully restrained laterally, the buckling resistance moment Mb of the section needs to be checked in comparison with the applied equivalent uniform moment M to ensure that
M = mMA ≤ Mb = pbSx
By reference to the bending moment diagram shown in Figure 5.25c, the critical unrestrained length will be BC where the maximum moment occurs.
The self-weight UDL is relatively insignificant and it is therefore satisfactory to consider the beam to be unloaded between restraints. Hence n is 1.0 and m is obtained from BS 5950 Table 18. We have
ß =smaller end moment M at C 286.59 larger end moment =
M at B 332.61= = 0.86
Therefore by interpolation from Table 18, m = 0.93. The maximum moment on length BC is MA = M at B = 332.61 kN m. Hence
M = mMA = 0.93 × 332.61 = 309.33 kN m The effective length LE of BC is 3.0 m.
By reference to Table 5.9, reproduced from the Steel Construction Institute design guide, a 457 × 191 × 74 kg/m UB has a buckling resistance moment of 355 kN m when n is 1 and LE is 3.0 m. Therefore let us check this section in bending, shear and deflection.
The relevant properties for the section from tables are as follows:
Plastic modulus Sx = 1660 cm3
D = 457.2 mm t = 9.1mm T = 14.5 mm d = 407.9 mm Section classification: plastic
Since T = 14.5 mm < 16 mm, py = 275 N/mm2.
Check the section for combined moment and shear as follows.
Maximum moment and coexistent shear at B
Ultimate shear at B is Fv = 108.77 kN; M = 332.61 kN m Shear capacity of section is Pv = 0.6 py t D
= 0.6 × 275 × 9.1 × 457.2 = 686 486 N
= 686 kN > 108.77 kN
This is satisfactory. Furthermore, 0.6 Pv = 0.6 × 686 = 412 kN. Therefore Fv = 108.77 kN < 0.6Pv = 412 kN
Hence the shear load is low and the moment capacity is as follows:
Mcx = pySx = 275 × 1660 × 103 = 456.5 × 106 N mm = 456.5 kN m > 332.61 kN m Maximum shear and coexistent moment at A
Ultimate shear at A is Fv = 112.97 kN; M = 0 Pv is again 686 kN > 112.97 kN
Buckling resistance
The lateral torsional buckling resistance has already been satisfied by selecting a section from Table 5.9 with a buckling resistance moment Mb greater than the equivalent uniform moment M. However, the method of calculating the buckling resistance moment in accordance with BS 5950 will be included here for reference.
The buckling resistance moment of the section is given by Mb = pbSx
The bending strength pb is obtained from Table 5.5 in relation to py and λLT. We have py = 275 N/mm2 and
λ LT = nuvλ
Now n = 1.0, and u = 0.876 from section tables. Next λ = LE/ry, where LE = 1.0 L in this instance from Table 5.6; L is the distance BC between restraints; and ry = 4.19 cm = 4.19 × 10 mm from section tables. Thus
LE 1.0 × 3000 λ = = 4.19 × 10 = 71.6
ry
Here x = 33.9 from section tables. Thus λ /x = 71.6/33.9 = 2.11, and so v = 0.95 by interpolation from Table 5.7.
Finally, therefore,
LT = nuvλ = 1.0 × 0.876 × 0.95 × 71.6 = 59.6
Using the values of py and LT, pb = 214 N/mm2 from Table 5.5. In conclusion, Mb = pbSx = 214 × 1660 × 103 = 355.2 × 106 N mm = 355.2 kN m > 309.33 kN m Thus M < Mb, and therefore the lateral torsional buckling resistance of the section is adequate.
Deflection
Since the loading on this beam is not symmetrical, the calculations needed to determine the actual deflection are quite complex. A simpler approach is to calcu-late the deflection due to an equivalent UDL and compare it with the permitted limit of span/360. If this proves that the section is adequate then there would be no need to resort to more exact calculations.
λ
λ
The deflection should be based upon the unfactored imposed loads alone. These and the resulting shear and bending moment diagrams are shown in Figure 5.26.
By equating the maximum bending moment of 129 kN m to the expression for the bending moment due to a UDL, an equivalent UDL can be calculated:
129 =WL 8
W 8 × 129 8 × 129
= L =
9 = 115 kN
This equivalent UDL of 115 kN may be substituted in the expression for the deflection of a simply supported beam:
Actual deflection δ a 5
5 WL3 115 × 103 × 90003
= = =384 EI 384 ×
205 × 103 × 33 400 × 104 15.94 mm Deflection limit δ p span 9000
=360 = 360 = 25 mm
Thus δ a < δ p, and the beam is satisfactory in deflection.
(a) Unfactored imposed load diagram
43 kN
7 kN
37 kN (b) Shear force diagram
+ 129 kNm
111 kNm
(c) Bending moment diagram
Figure 5.26 Beam diagrams for unfactored imposed loads
A Ra = 43 kN
50 kN
3 m B
30 kN
C 3 m 9 m
3 m D
Rd 37 kN
+
–
Web buckling and bearing
The web buckling and bearing requirements are not critical and therefore the calculations for these will be omitted.
Conclusion
That completes the check on the section, which has been shown to be adequate in bending, shear and deflection. Thus:
Adopt 457 × 191 × 74 kg/m UB.
5.11 Fabricated beams
In situations where standard rolled sections are found to be inadequate, consideration should be given to the following fabricated alternatives.Compound beams
The strength of standard rolled sections can be increased by the addition of reinforcing plates welded to the flanges. Beams strengthened in this way are called compound beams. Examples are shown in Figure 5.27.
Welded flange plate
Basic rolled section
Welded flange plate
Basic rolled section
Welded flange plate
Basic rolled section
Welded flange plate Welded flange plate
Figure 5.27 Examples of compound beams
Castellated beams
Standard rolled sections can be converted by cutting and welding into much deeper sections known as castellated beams. They offer a relatively simple method of increasing the strength of a section without increasing its weight.
To form a castellated beam, the basic rolled section is first flame cut along its web to a prescribed profile as shown in Figure 5.28a. Then the resulting two halves are rejoined by welding to form the castellated beam shown in Figure 5.28 b. The finished section is stronger in bending than the original but the shear strength is less. However, this usually only affects heavily loaded short span beams, and may be overcome where necessary by welding fitted plates into the end castellations as shown in Figure 5.28c.
Weld Trimmed to required length
1.5 d
(b) Two-halves re-joined to form castellated beam
Fitted plate welded into end castellation if necessary
(c) Method of catering for shear
Figure 5.28 Castellated beams
Flange angles Tongue plate
Web plate
Tongue plate Web plate
Flange angles Flange plate
Flange plate
Flange plate
Web plate
Flange plate
Flange plate
Flange plate
Figure 5.29 Examples of plate girders
Line of cut
d 60° 0.5 d
(a) Web of basic rolled section cut to prescribed profile
5.12 Columns
Plate girders
Plate girders are used occasionally in buildings where heavy loads or long spans dictate, but more often they are used for bridges. They are formed from steel plates, sometimes in conjunction with angles, which are welded or bolted together to form I-sections. Three of the most common forms are illustrated in Figure 5.29.
Whilst plate girders can theoretically be made to any size, their depth for practical reasons should usually be between span/8 and span/12.
Lattice girders
Lattice girders are a framework of individual members bolted or welded together to form an open web beam. Two types of lattice girder commonly encountered are illustrated in Figure 5.30; they are the N-girder and the Warren girder. In comparison with the structural behaviour of beams, the top and bottom booms of a lattice girder resist bending and the internal members resist shear.
Top compression boom Internals
Bottom tension boom N-girder
Internals Top compression boom
Bottom tension boom Warren girder
Figure 5.30 Examples of lattice girders
Generally their economical depth is between span/10 and span/15. Ex-ceptions are short span heavily loaded girders, for which the depth may equal span/6, and long span lightly loaded roof girders, for which a depth of span/20 may suffice.
A steel column may be subject to direct compression alone, where the load is applied axially, or subject to a combination of compressive loading and bending due to the load being applied eccentrically to the member
axes. It may also be subject to horizontal bending induced by lateral wind loading. However, the effect of wind loading on individual structural elements is not being considered in this manual.
Guidance for the design of axially loaded columns and axially loaded columns with moments is given in BS 5950 Part 1. The procedure for dealing with columns subject to axial load alone is first explained. This is then extended to include the interaction between compression and bend-ing. Separate guidance is also given for the design of concrete cased col-umns and baseplates for colcol-umns.
The design of steel columns in this manual will therefore be considered under the following headings:
(a) Axially loaded columns
(b) Axially loaded columns with moments (c) Cased columns
(d) Column base plates.